0:00:00.600,0:00:05.160
We're told that glucose reacts[br]with oxygen to give carbon

0:00:05.160,0:00:07.240
dioxide and water.

0:00:07.240,0:00:11.190
What mass of oxygen, in grams,[br]is required to complete the

0:00:11.190,0:00:15.950
reaction of 25 grams[br]of glucose?

0:00:15.950,0:00:18.990
And they also want to know what[br]masses of carbon dioxide

0:00:18.990,0:00:20.405
and water are formed?

0:00:20.405,0:00:23.160
Well let's first just[br]write the reaction.

0:00:23.160,0:00:27.540
So they're saying glucose reacts[br]with oxygen to give

0:00:27.540,0:00:30.030
carbon dioxide and water.

0:00:30.030,0:00:41.560
So glucose, C6H12O6, reacts with[br]oxygen in its molecular

0:00:41.560,0:00:43.595
form-- there's two[br]atoms there.

0:00:43.595,0:00:46.630


0:00:46.630,0:00:51.135
The products are carbon[br]dioxide and water.

0:00:51.135,0:00:54.200


0:00:54.200,0:00:56.450
And you might already[br]recognize this.

0:00:56.450,0:00:58.250
This is a stoichiometry[br]problem.

0:00:58.250,0:01:01.260
They're saying, hey, we have[br]25 grams of glucose.

0:01:01.260,0:01:03.370
How much oxygen is required[br]to completely

0:01:03.370,0:01:04.780
react with that glucose?

0:01:04.780,0:01:07.630
And then how much carbon dioxide[br]and how much water is

0:01:07.630,0:01:09.850
going to be produced in grams?

0:01:09.850,0:01:12.720
That's what stoichiometry[br]problems are all about.

0:01:12.720,0:01:16.250
And if you remember from the[br]last video, the first thing

0:01:16.250,0:01:18.210
you should always do[br]is make sure that

0:01:18.210,0:01:19.550
your equation is balanced.

0:01:19.550,0:01:21.600
So let's make sure[br]it's balanced.

0:01:21.600,0:01:24.960
On the left hand side-- and you[br]always want to do the most

0:01:24.960,0:01:27.830
complicated molecules first,[br]and then do the simplest

0:01:27.830,0:01:30.550
molecules last, because those[br]are the easiest ones to

0:01:30.550,0:01:32.010
balance out.

0:01:32.010,0:01:37.010
So on the left hand side, here,[br]I have a 6 carbons.

0:01:37.010,0:01:38.490
On the entire left hand side.

0:01:38.490,0:01:41.050
On the right hand side,[br]I only have 1 carbon.

0:01:41.050,0:01:45.110
So let me multiply this[br]over here by 6.

0:01:45.110,0:01:49.380
And now I have 6 carbons on both[br]sides of this equation.

0:01:49.380,0:01:50.490
Let's move to the hydrogens.

0:01:50.490,0:01:52.880
I have 12 hydrogens[br]on the left hand

0:01:52.880,0:01:54.060
side of this equation.

0:01:54.060,0:01:56.740
The 12 are all sitting right[br]there in the glucose.

0:01:56.740,0:01:58.850
How many do I have on[br]the right hand side?

0:01:58.850,0:02:00.910
Well I only have 2 hydrogens.

0:02:00.910,0:02:04.120
So let me multiply[br]that times 6.

0:02:04.120,0:02:06.210
And I didn't mess with the[br]carbons at all, so that

0:02:06.210,0:02:09.650
shouldn't change anything-- so[br]now I have 12 hydrogens on

0:02:09.650,0:02:10.759
both sides of this equation.

0:02:10.759,0:02:12.850
12 here, 12 there.

0:02:12.850,0:02:14.690
And now I can just balance[br]out the oxygen.

0:02:14.690,0:02:17.030
I saved that for last because[br]I just have the oxygen

0:02:17.030,0:02:17.650
molecule here.

0:02:17.650,0:02:19.450
That's the easiest[br]one to balance.

0:02:19.450,0:02:21.470
So how may oxygens do I have[br]on the right hand side?

0:02:21.470,0:02:23.380
I have 6 times 2 here.

0:02:23.380,0:02:25.760
I have 12 oxygens there.

0:02:25.760,0:02:28.660
And then I have another[br]6 oxygens over here.

0:02:28.660,0:02:30.600
So plus 6 oxygens.

0:02:30.600,0:02:35.310
I have 18 oxygens on the right[br]hand side of my equation.

0:02:35.310,0:02:37.400
So I need to have 18[br]oxygens on the left

0:02:37.400,0:02:38.870
hand side of my equation.

0:02:38.870,0:02:40.660
How many do I have right now?

0:02:40.660,0:02:45.500
I have 6 oxygens over here.

0:02:45.500,0:02:47.800
So I'm going to need[br]12 over here.

0:02:47.800,0:02:47.990
Right?

0:02:47.990,0:02:49.460
This is the last thing[br]I want to mess with.

0:02:49.460,0:02:51.790
I don't want to put a[br]coefficient out here.

0:02:51.790,0:02:53.320
That'll change everything[br]else.

0:02:53.320,0:02:54.630
I just want to put a[br]coefficient here.

0:02:54.630,0:02:56.610
That'll make everything[br]balance out.

0:02:56.610,0:02:58.450
I have 18 on the right[br]hand side.

0:02:58.450,0:03:00.440
I already have 6 here.

0:03:00.440,0:03:02.580
I want to have 12 right here.

0:03:02.580,0:03:05.150
So let me multiply[br]this times 6.

0:03:05.150,0:03:06.470
And now everything[br]should work out.

0:03:06.470,0:03:08.580
I have 6 carbons[br]on both sides.

0:03:08.580,0:03:11.640
I have 12 hydrogens[br]on both sides.

0:03:11.640,0:03:13.980
And I have 18 oxygens.

0:03:13.980,0:03:16.100
Six here, 12 here.

0:03:16.100,0:03:17.170
12 here, 6 here.

0:03:17.170,0:03:20.870
I have 18 oxygens on both[br]sides of this equation.

0:03:20.870,0:03:23.260
Now, the next thing we're going[br]to want to do is figure

0:03:23.260,0:03:26.290
out how many moles of the[br]reactants that they're telling

0:03:26.290,0:03:29.150
us about that we have. So[br]they're telling us that we

0:03:29.150,0:03:37.040
have 25.0 grams of glucose.

0:03:37.040,0:03:40.780
So let's figure out how many[br]moles per gram, or how many

0:03:40.780,0:03:43.690
grams per mole, there are[br]of a glucose molecule.

0:03:43.690,0:03:45.440
And since everything here is[br]dealing with carbons, and

0:03:45.440,0:03:49.120
oxygens, and hydrogens, let's[br]look up the atomic weights of

0:03:49.120,0:03:50.570
all of them.

0:03:50.570,0:03:53.750
So carbons, hydrogens, and[br]oxygens are pretty common.

0:03:53.750,0:03:55.280
So at some point you might[br]want to memorize

0:03:55.280,0:03:56.050
their atomic weights.

0:03:56.050,0:04:00.230
And I want to give proper credit[br]to the person whose

0:04:00.230,0:04:01.790
periodic table I'm using.

0:04:01.790,0:04:03.310
Le Van Han Cedric.

0:04:03.310,0:04:05.030
I got this off of Wikimedia.

0:04:05.030,0:04:07.020
It's a creative commons[br]attribution license.

0:04:07.020,0:04:09.330
So I want to make sure I[br]attribute the person who made

0:04:09.330,0:04:11.330
the periodic table.

0:04:11.330,0:04:12.690
But we have oxygen.

0:04:12.690,0:04:17.970
It has an atomic weight[br]of 15.9999.

0:04:17.970,0:04:20.769
Usually it's given as 16.

0:04:20.769,0:04:21.990
But I'll just write[br]it like that.

0:04:21.990,0:04:28.493
So oxygen is 15.999.

0:04:28.493,0:04:30.020
And we're going to[br]have to figure

0:04:30.020,0:04:32.810
out carbon, and hydrogen.

0:04:32.810,0:04:36.190


0:04:36.190,0:04:38.760
Let's go back to the[br]periodic table.

0:04:38.760,0:04:45.960
We have carbon, has an atomic[br]weight of 12.011.

0:04:45.960,0:04:50.186
And hydrogen is 1.0079.

0:04:50.186,0:04:55.330
So let's do the carbon,[br]12.011.

0:04:55.330,0:04:59.350
So the carbon is 12.011.

0:04:59.350,0:05:02.900
And then the hydrogen-- remember[br]atomic weights are

0:05:02.900,0:05:05.380
all the weighted average[br]of all of the isotopes

0:05:05.380,0:05:10.050
on earth-- is 1.0079.

0:05:10.050,0:05:14.180
1.00-- what was it?

0:05:14.180,0:05:15.380
Was it one zero or two zeroes?

0:05:15.380,0:05:17.060
It was two zeroes.

0:05:17.060,0:05:20.560
1.0079.

0:05:20.560,0:05:26.110
So what's the atomic weight of[br]glucose, given all of that?

0:05:26.110,0:05:28.230
So what's the atomic[br]weight of glucose?

0:05:28.230,0:05:30.500
Let me scroll down[br]a little bit.

0:05:30.500,0:05:37.710
So the atomic weight of glucose,[br]C6H12O6, it's going

0:05:37.710,0:05:42.580
to be equal to 12.011 times 6.

0:05:42.580,0:05:46.110
6 times 12.011.

0:05:46.110,0:05:48.900
Plus 12 times hydrogen.

0:05:48.900,0:05:50.900
Just to keep things simple,[br]actually, let me just make

0:05:50.900,0:05:53.360
this into 16.

0:05:53.360,0:05:56.630
let me make this into 12, and[br]let me make this into 1.

0:05:56.630,0:05:59.110
That's going to make our[br]math a lot easier.

0:05:59.110,0:06:01.710
So let me just do[br]it like that.

0:06:01.710,0:06:04.520
Let me clear this.

0:06:04.520,0:06:10.570
So the atomic weight of glucose[br]is 6 times the atomic

0:06:10.570,0:06:11.770
weight of carbon.

0:06:11.770,0:06:13.960
Six times 12.

0:06:13.960,0:06:15.450
Six times 12.

0:06:15.450,0:06:17.720
Plus 12 times the atomic[br]weight of hydrogen.

0:06:17.720,0:06:22.380
Plus one times 12, or maybe I[br]have to write 12 times one,

0:06:22.380,0:06:23.840
just to be consistent.

0:06:23.840,0:06:25.100
12 times one.

0:06:25.100,0:06:29.320
Plus 6 times the atomic[br]weight of oxygen.

0:06:29.320,0:06:31.320
Plus 6 times 16.

0:06:31.320,0:06:31.930
So what is this?

0:06:31.930,0:06:35.830
This is equal to 72 plus 12.

0:06:35.830,0:06:41.870
Plus, 6 times 16[br]is 60 plus 36.

0:06:41.870,0:06:45.490
It's 96.

0:06:45.490,0:06:50.020
And that is-- I'll get the[br]calculator out-- 72.

0:06:50.020,0:06:58.530
Plus 12, plus 96,[br]is equal to 180.

0:06:58.530,0:07:03.870
So the atomic weight of glucose[br]is equal to 180.

0:07:03.870,0:07:08.940
Which tells us, that let me[br]start doing the problem.

0:07:08.940,0:07:13.790
We have 25.0 grams of glucose.

0:07:13.790,0:07:19.780
So I'll just write[br]grams of C6H12O6.

0:07:19.780,0:07:23.490
We want to write this in terms[br]of moles of glucose.

0:07:23.490,0:07:28.410
So we want to cancel out the[br]grams. So we want the grams in

0:07:28.410,0:07:29.640
the denominator.

0:07:29.640,0:07:29.890
Right?

0:07:29.890,0:07:30.730
Here it's in the numerator.

0:07:30.730,0:07:34.750
If we divide by that unit, we're[br]going to cancel it out.

0:07:34.750,0:07:42.390
So we want the grams of C6H12O6,[br]or the grams of

0:07:42.390,0:07:43.730
glucose in the denomenator.

0:07:43.730,0:07:50.180
And we want the moles of glucose[br]in the numerator.

0:07:50.180,0:07:54.520
Because then when we perform[br]this calculation, that will

0:07:54.520,0:07:57.270
cancel with that, and we will[br]be left with moles.

0:07:57.270,0:07:59.070
So how many grams per mole?

0:07:59.070,0:08:00.290
Well why don't we just[br]figure it out?

0:08:00.290,0:08:02.800
The atomic weight is a 180.

0:08:02.800,0:08:03.970
So it's 180.

0:08:03.970,0:08:08.870
If we have Avagadro's number of[br]these molecules, it's going

0:08:08.870,0:08:15.140
to have a mass, I should say,[br]of 180 grams per mole.

0:08:15.140,0:08:19.150
That's the information we got[br]by figuring out glucose's

0:08:19.150,0:08:19.900
atomic weight.

0:08:19.900,0:08:22.010
So let's just perform[br]this calculation.

0:08:22.010,0:08:26.220
These units cancel out[br]with those units.

0:08:26.220,0:08:29.940
And so we just take 25 times[br]one, divided by 180.

0:08:29.940,0:08:34.018
So this is equal[br]to 25 over 180.

0:08:34.018,0:08:35.950
And all were left[br]with in units is

0:08:35.950,0:08:42.806
moles C6H12O6, or glucose.

0:08:42.806,0:08:46.330
And what is 25 five[br]divided by 180?

0:08:46.330,0:08:49.820
I should say 25.0 just so we[br]know that we have three

0:08:49.820,0:08:52.840
significant digits here.

0:08:52.840,0:08:54.960
And I don't want to[br]be too particular

0:08:54.960,0:08:55.870
about significant digits.

0:08:55.870,0:08:58.080
Sometimes I'm a little bit[br]loosey goosey about it.

0:08:58.080,0:09:00.370
And I was already loosey goosey[br]about the topic weight.

0:09:00.370,0:09:02.160
But we'll try to be someplace[br]in the ballpark.

0:09:02.160,0:09:12.420
So 25 divided by 180[br]is equal to 0.139.

0:09:12.420,0:09:20.060
So we have 0.139 moles of-- I'm[br]tired of writing, I'm just

0:09:20.060,0:09:23.100
going to write-- moles[br]of glucose.

0:09:23.100,0:09:28.180
Now, we want to figure out how[br]many moles of oxygen-- or,

0:09:28.180,0:09:31.160
well, the first step we want[br]to figure out is for every

0:09:31.160,0:09:34.150
mole of glucose, how many[br]moles on oxygen?

0:09:34.150,0:09:34.860
And we know that.

0:09:34.860,0:09:36.400
It's 1 to 6.

0:09:36.400,0:09:37.660
So let's write that down.

0:09:37.660,0:09:40.216
And that's going to tell us how[br]many moles of oxygen we're

0:09:40.216,0:09:41.060
going to need.

0:09:41.060,0:09:43.810
And, remember, we want to get[br]rid of these moles of glucose.

0:09:43.810,0:09:45.565
So we want to write that[br]in the denominator.

0:09:45.565,0:09:49.870


0:09:49.870,0:09:56.490
So moles of glucose required.

0:09:56.490,0:10:00.600
For every 1 mole of glucose[br]required, we figured out when

0:10:00.600,0:10:03.680
we balanced out the equation,[br]for every mole of glucose, we

0:10:03.680,0:10:06.510
need 6 moles of oxygen.

0:10:06.510,0:10:13.060
We need 6 moles of[br]O2 required.

0:10:13.060,0:10:16.360
And then what do we get here?

0:10:16.360,0:10:20.190
Well we have the moles of[br]glucose canceling out with the

0:10:20.190,0:10:21.750
moles of glucose.

0:10:21.750,0:10:26.180
And then we multiply[br]the 6 times 0.139.

0:10:26.180,0:10:32.850
So let's just multiply this[br]times 6 is equal to 0.833.

0:10:32.850,0:10:38.770
So we need 0-- that's[br]too fat-- 0.833

0:10:38.770,0:10:43.330
moles of oxygen required.

0:10:43.330,0:10:47.720
Now, we're almost there, we know[br]how many moles of oxygen

0:10:47.720,0:10:49.610
are required, molecular[br]oxygen.

0:10:49.610,0:10:52.130
Now we just have to figure out[br]how many grams that is.

0:10:52.130,0:10:53.820
So let me just write[br]it over here.

0:10:53.820,0:11:01.850
We have 0.833 moles of molecular[br]oxygen are required.

0:11:01.850,0:11:04.010
We're going to multiply that.

0:11:04.010,0:11:07.470
Remember, we want to get[br]this into grams now.

0:11:07.470,0:11:10.570
So let's try the moles of[br]O2 in the denomenator.

0:11:10.570,0:11:13.350


0:11:13.350,0:11:17.090
And we're going to put grams[br]of O2 in the numerator.

0:11:17.090,0:11:21.300
So how many grams of O2[br]are there per mole?

0:11:21.300,0:11:24.980
Well, we know the atomic[br]weight oxygen is 16.

0:11:24.980,0:11:27.460
So the atomic weight of[br]molecular oxygen, where you

0:11:27.460,0:11:33.520
have two of these, is equal[br]to 16 times two.

0:11:33.520,0:11:36.400
Which is equal to 32.

0:11:36.400,0:11:42.640
So it's 32 grams per mole.

0:11:42.640,0:11:46.400
So when we perform the[br]calculation, the moles of O2

0:11:46.400,0:11:48.950
in the numerator cancel out[br]with moles of O2 in the

0:11:48.950,0:11:49.680
denomenator.

0:11:49.680,0:11:59.750
And we're just going to have[br]0.833 times 32 is equal to

0:11:59.750,0:12:02.940
26-- I'll just say, seven.

0:12:02.940,0:12:09.570
So this is equal to 26.7[br]grams of O2 required.

0:12:09.570,0:12:13.650
Grams of O2, and I'll[br]write required in

0:12:13.650,0:12:15.650
that same exact collar.

0:12:15.650,0:12:17.940
So we finished one part[br]of the question.

0:12:17.940,0:12:20.830
We have two more parts left.

0:12:20.830,0:12:25.180
If we go back to the problem, it[br]asks us, we already figured

0:12:25.180,0:12:27.680
out, what mass of oxygen[br]is required for

0:12:27.680,0:12:28.470
the complete reaction?

0:12:28.470,0:12:31.500
We did that part, so we can[br]write a little check there.

0:12:31.500,0:12:32.500
We did that first part.

0:12:32.500,0:12:36.690
But they also ask, what masses[br]of carbon dioxide and water

0:12:36.690,0:12:37.490
are formed?

0:12:37.490,0:12:41.410
So now we need to figure out[br]carbon dioxide and water.

0:12:41.410,0:12:43.800
And so we go back here.

0:12:43.800,0:12:47.930
We know we have 0.139[br]moles of glucose.

0:12:47.930,0:12:50.550
We figured that out[br]in the first part.

0:12:50.550,0:12:58.350
So 0.139 moles of glucose.

0:12:58.350,0:13:05.140
And let's do the carbon dioxide[br]first. We know for

0:13:05.140,0:13:09.940
every mole of glucose, we're[br]going to produce 6 moles of

0:13:09.940,0:13:10.770
carbon dioxide.

0:13:10.770,0:13:11.670
We see that right there.

0:13:11.670,0:13:13.560
There's a one out[br]here implicitly.

0:13:13.560,0:13:17.380
So for every mole of glucose,[br]you're going to produce 6

0:13:17.380,0:13:19.290
moles of carbon dioxide.

0:13:19.290,0:13:21.130
So let's write that.

0:13:21.130,0:13:29.040
So for every one mole of[br]glucose, you're going to have

0:13:29.040,0:13:34.660
6 moles of CO2 produced.

0:13:34.660,0:13:36.780
We just got that directly[br]from the equation.

0:13:36.780,0:13:39.710
And I wrote it this way, instead[br]of one over the 6,

0:13:39.710,0:13:41.980
because I wanted it to cancel[br]out with the moles of glucose.

0:13:41.980,0:13:45.050
Moles of glucose in the[br]numerator, moles of glucose in

0:13:45.050,0:13:46.920
the denominator.

0:13:46.920,0:13:48.870
So that canceled[br]out with that.

0:13:48.870,0:13:55.960
And this is going to be[br]equal to 6 times 0.139

0:13:55.960,0:14:01.840
moles of CO2 produced.

0:14:01.840,0:14:04.810


0:14:04.810,0:14:05.710
And we know what that is.

0:14:05.710,0:14:09.040
We already multiplied 6[br]times 0.139 before.

0:14:09.040,0:14:17.300
This is going to be equal to[br]0.833 moles of CO2 produced.

0:14:17.300,0:14:19.790
We've already done[br]that calculation.

0:14:19.790,0:14:22.960
And actually we could even--[br]well, I won't skip too many

0:14:22.960,0:14:29.120
steps-- but we see over here,[br]for every one mole of glucose,

0:14:29.120,0:14:32.890
6 moles of this, 6 moles of[br]this, and 6 moles of that.

0:14:32.890,0:14:38.850
So that's why we had the[br]0.833 moles of oxygen.

0:14:38.850,0:14:43.570
And we're also getting the 0.833[br]moles of carbon dioxide.

0:14:43.570,0:14:46.300
And, if we did the exact same[br]thing, we've performed the

0:14:46.300,0:14:47.480
exact same calculation.

0:14:47.480,0:14:49.515
For every mole of this[br]used, you have a

0:14:49.515,0:14:50.580
mole of that produced.

0:14:50.580,0:14:53.180
Or every 6 of this, you have[br]6 of that, 6 of that.

0:14:53.180,0:14:55.510
So you're also going to have,[br]by the very same logic, I

0:14:55.510,0:14:57.470
mean, you could do this[br]again with the water.

0:14:57.470,0:15:06.980
You're also going to have 0.833[br]moles of water produced.

0:15:06.980,0:15:07.270
Right?

0:15:07.270,0:15:09.330
For every mole of carbon[br]dioxide, you

0:15:09.330,0:15:10.100
have a mole of water.

0:15:10.100,0:15:12.620
Here it's 6 for 6, but[br]it's a 1:1 ratio.

0:15:12.620,0:15:14.530
You could think about[br]it like that.

0:15:14.530,0:15:18.710
Now let's figure out how[br]many grams this is.

0:15:18.710,0:15:20.920
We have the same number of moles[br]of carbon dioxide and

0:15:20.920,0:15:23.330
the same number of[br]moles of water.

0:15:23.330,0:15:26.400
But they're going to have[br]different number of grams. So

0:15:26.400,0:15:30.340
the actual number of molecules[br]are the same, but the actual

0:15:30.340,0:15:32.480
mass of those molecules are[br]going to be different.

0:15:32.480,0:15:39.950
So what's carbon dioxide's[br]atomic weight?

0:15:39.950,0:15:41.410
Just to remind ourselves.

0:15:41.410,0:15:45.520
Carbon has atomic weight[br]of 12, oxygen of 16.

0:15:45.520,0:15:50.810
So it's going to be 12 plus 16[br]times two, which is equal to

0:15:50.810,0:15:55.510
12 plus 32, which[br]is equal to 44.

0:15:55.510,0:16:06.370
So if we are starting off with[br]0.833 moles of carbon dioxide

0:16:06.370,0:16:09.550
that are produced, we want to[br]figure out how many grams per

0:16:09.550,0:16:12.320
mole-- we're going to put the[br]moles of carbon dioxide in the

0:16:12.320,0:16:15.220
denominator because we want[br]it to cancel with that.

0:16:15.220,0:16:18.360
And then we have the grams[br]of carbon dioxide.

0:16:18.360,0:16:21.090
We know carbon dioxide's[br]atomic weight is 44.

0:16:21.090,0:16:24.510
So that means there are 44[br]grams. If we have a mole of

0:16:24.510,0:16:27.620
carbon dioxide, if we have[br]6.022 times 10 to the 23

0:16:27.620,0:16:29.680
carbon dioxide molecules.

0:16:29.680,0:16:34.590
So this is going to[br]be 44 times 0.833

0:16:34.590,0:16:38.130
is-- what do we get?

0:16:38.130,0:16:46.500
0.833 times 44 is equal to 36[br]point-- let's just say, 7.

0:16:46.500,0:16:56.220
36.7 grams of CO2 are going[br]to be produced.

0:16:56.220,0:16:57.530
And then we're almost done.

0:16:57.530,0:17:00.380
We just have to figure out how[br]many grams of water are going

0:17:00.380,0:17:01.040
to be produced.

0:17:01.040,0:17:02.590
We know it's the same number[br]of moles, but how

0:17:02.590,0:17:04.329
many grams is that?

0:17:04.329,0:17:11.470
So water's atomic weight is[br]going to be 2 times 1, right?

0:17:11.470,0:17:13.460
Because we have to[br]2 hydrogens.

0:17:13.460,0:17:15.839
Plus 16.

0:17:15.839,0:17:18.500
Which is equal to 18.

0:17:18.500,0:17:26.160
So if we're starting off with[br]0.833 moles, or if we're

0:17:26.160,0:17:33.680
producing 0.833 moles of water,[br]we just multiply that

0:17:33.680,0:17:36.920
times how many grams[br]per mole of water?

0:17:36.920,0:17:41.200
So 1 mole of H2O-- once again[br]we want this in the

0:17:41.200,0:17:45.110
denominator so it cancels out[br]with this in the numerator.

0:17:45.110,0:17:51.990
1 mole of H20 has a mass of 18[br]grams. Or 18 grams of H20 for

0:17:51.990,0:17:53.700
every mole of H20.

0:17:53.700,0:17:59.510
So this cancels with that and[br]we get-- get the calculator

0:17:59.510,0:18:13.180
out-- 18 times 0.833[br]is equal to 14.994.

0:18:13.180,0:18:16.630
So we can just round[br]that to 15.0.

0:18:16.630,0:18:25.230
So we have 15.0 grams[br]of water produced.

0:18:25.230,0:18:27.200
And we're done!

0:18:27.200,0:18:28.520
We are done.

0:18:28.520,0:18:34.910
We figured out that if we start[br]off with that 25 grams

0:18:34.910,0:18:37.180
of glucose, like they told us[br]at the beginning of the

0:18:37.180,0:18:42.600
problem, we're going to require[br]26.7 grams of oxygen

0:18:42.600,0:18:46.740
to react with it, we're going[br]to produce 36.7 grams of

0:18:46.740,0:18:52.180
carbon dioxide, and[br]15 grams of water.